This MCQ module is based on: LCM and Prime Factorisation
TOPIC 8 OF 23
LCM and Prime Factorisation
🎓 Class 7
Mathematics
CBSE
Theory
Ch 3 — Factors and Multiples
⏱ ~35 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📐 Maths Assessment
▲
This mathematics assessment will be based on: LCM and Prime Factorisation
Targeting Class 7 level in General Mathematics, with Basic difficulty.
Upload images, PDFs, or Word documents to include their content in assessment generation.
3.3 Common Multiples (LCM)
Kabamali and Asangla are sisters. Kabamali visits her native village every 14 days and Asangla every 10 days. They left together for the village today. When will they next leave together?
Let's find out! Kabamali will return on days that are multiples of 14: 14, 28, 42, 56, 70, 84, 98, 112, 126, 140, ... days.
Asangla will arrive at the sweet shop on multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, ... days.
These are multiples of 10. When will Kabamali eat fresh sweets again? It will happen on days common to the sequences of multiples above. It can be seen that this will first happen after 70 days. Notice that, here too, 70 is the lowest among all the common multiples of 7 and 10.
For both these problems the solution was the lowest common multiple? of two or more given numbers, known in the lowest (or smallest or least) of their common multiples.
Definition — LCM
The Lowest Common Multiple (LCM) of two or more given numbers is the smallest of their common multiples. For 7 and 10, LCM = 70.
🔵 Do you remember the 'Idli–Vada' game from Grade 6 chapter on Prime Time? Two numbers are chosen and whenever players come to their multiples, 'idli' or 'vada' should be called out depending on whose multiple the number is. If the number happens to be a common multiple, the two numbers corresponding to 'idli' and 'vada' are given. Find the first number for which 'idli–vada' will be called out:
(a) 4 and 6 (b) 7 and 11 (c) 14 and 55 (d) 15 and 55
Answers: (a) 12, (b) 77, (c) 770, (d) 165.
(a) 4 and 6 (b) 7 and 11 (c) 14 and 55 (d) 15 and 55
Answers: (a) 12, (b) 77, (c) 770, (d) 165.
Is the answer always the LCM of the two numbers? Explain. — Yes. The first number where both multiples meet is, by definition, the LCM.
As in the case of the HCF, the process of finding the LCM by listing down the multiples may get tedious for larger numbers, as you would have seen for questions (c) and (d) above. Prime factorisation can simplify the process of finding the LCM as well.
3.4 Finding LCM from Prime Factorisation
We have seen that every factor of a number is formed by taking a subpart of its prime factorisation. We used this fact to come up with a method to find the HCF of two numbers. In a similar manner, we can come up with a method to find the LCM.
We begin by comparing the prime factorisations of a number and a multiple of that number. For example, let us take 36 and its multiple \(648 (= 36 \times 18)\).
Prime factorisation of 36 = 2 × 2 × 3 × 3.
Prime factorisation of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 (\(= 2^3 \times 3^4\)).
All the prime factors of 36 appear in the prime factorisation of 648. This is true in general: if \(a\) is a factor of \(b\), then all the primes of \(a\) (with their powers) appear in \(b\).
So, to find a common multiple of two numbers, we just need a number whose prime factorisation contains all the primes of both numbers (with at least the highest power each appears in either).
Rule for LCM
To find the LCM of two numbers using prime factorisation:
- Write prime factorisations of both numbers.
- For each prime that appears in either, take the highest power it appears in.
- Multiply these together.
Example: LCM(84, 180)
Observe that \(84 = 2 \times 2 \times 3 \times 7\) and \(180 = 2 \times 2 \times 3 \times 3 \times 5\), so the LCM \(= 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 1260\).
Division ladder method: HCF uses only common prime factors; LCM continues until both reduce to 1.
Figure it Out — Finding the HCF and LCM
Q1. Find the HCF and LCM using prime factorisation: (a) 300, 150 (b) 630, 770
(a) 300 = 2²×3×5²; 150 = 2×3×5². HCF = 2×3×5² = 150. LCM = 2²×3×5² = 300.
(b) 630 = 2×3²×5×7; 770 = 2×5×7×11. HCF = 2×5×7 = 70. LCM = 2×3²×5×7×11 = 6930.
(b) 630 = 2×3²×5×7; 770 = 2×5×7×11. HCF = 2×5×7 = 70. LCM = 2×3²×5×7×11 = 6930.
Why are these the LCMs?
Hint: Will the product of the factors marked as the LCM of 300 and 150 contain the prime factorisations of both 300 and 150? Is this the smallest such number? Guna says: "I found a better way to factorise for HCF/LCM. This is better than what we taught in class." For the numbers 300 and 150, I can first directly divide both numbers by 50. The HCF will be 50 × something.
This is similar to the procedure for prime factorisation. At each step, the two numbers are divided by a common prime factor, and the two quotients are written down in the next row. This continues till we get two numbers that do not have any common prime factors.
Interactive: HCF and LCM Explorer
L3 ApplyTry it: Enter two numbers (between 2 and 999).
HCF(84, 180) = 12, LCM(84, 180) = 1260. Verification: HCF × LCM = 12 × 1260 = 15120 = 84 × 180 ✓
HCF × LCM = Product of the two numbers
There is a delightful connection: for any two positive integers \(a\) and \(b\):
HCF(\(a, b\)) × LCM(\(a, b\)) = \(a \times b\).
This is because every prime factor appearing in either number is split between the HCF (minimum power) and the LCM (maximum power), and together they account for all copies of that prime in \(a \times b\).
Activity: Traffic-Light LCM
L3 ApplyMaterials: Three coloured LEDs or marked cards (red every 6 s, green every 8 s, yellow every 10 s).
Predict: If all three blink together now, after how many seconds will all three blink together again?
- Set up three 'blinks' with periods 6, 8, and 10 seconds.
- List the times each one blinks from 0 s onwards.
- Find the first time they coincide again.
- Verify using prime factorisation: 6 = 2×3, 8 = 2³, 10 = 2×5. LCM = 2³ × 3 × 5 = 120 s.
All three blink together every 120 seconds = 2 minutes. This is the LCM.
Competency-Based Questions
Scenario: In a relay run, three runners take 12 s, 18 s, and 24 s respectively to complete one lap. They start together at the same point at time 0. The coach wants to know when they all cross the start line together again.
Q1. After how many seconds do all three runners cross the start line together for the first time?
L3 Apply12 = 2²×3; 18 = 2×3²; 24 = 2³×3. LCM = 2³×3² = 72 seconds.
Q2. How many laps has each runner completed at that moment? Analyse whether the slowest runner completes a whole number of laps.
L4 AnalyseRunner 1: 72 ÷ 12 = 6 laps. Runner 2: 72 ÷ 18 = 4 laps. Runner 3: 72 ÷ 24 = 3 laps. Yes — each runs a whole number of laps. That is precisely the defining property of the LCM.
Q3. Evaluate the coach's claim that "If we add a fourth runner who takes 30 s per lap, they will all meet again at 72 s." Justify mathematically.
L5 EvaluateClaim is false. 30 does not divide 72 (72 ÷ 30 = 2.4). The new LCM = LCM(72, 30). 72 = 2³×3², 30 = 2×3×5 → LCM = 2³×3²×5 = 360 s. So all four runners meet only every 360 s.
Q4. Design lap times (in whole seconds, each between 5 and 40) for three runners so that they all meet every 60 seconds exactly, and no earlier. Justify.
L6 CreateSample: 20, 12, 15 seconds. 20 = 2²×5, 12 = 2²×3, 15 = 3×5. LCM = 2²×3×5 = 60. Each divides 60 and the LCM of all three is 60. Many valid answers exist — the requirement is that each number divides 60 and their LCM equals 60.
Assertion–Reason Questions
A: LCM(6, 9) = 18.
R: LCM of two numbers is the highest power of each prime appearing in either number, multiplied together.
R: LCM of two numbers is the highest power of each prime appearing in either number, multiplied together.
Answer: (a). 6 = 2×3, 9 = 3². Highest power: 2¹ × 3² = 18. Both true and R explains A.
A: For any two numbers \(a\) and \(b\), \(\text{HCF} \times \text{LCM} = a \times b\).
R: Each prime factor's contribution to HCF (minimum power) and LCM (maximum power) together equals its total power in \(a \times b\).
R: Each prime factor's contribution to HCF (minimum power) and LCM (maximum power) together equals its total power in \(a \times b\).
Answer: (a). True for any two positive integers. R explains why — minimum + maximum powers sum to the two original powers.
A: If HCF(a, b) = 1, then LCM(a, b) = a × b.
R: Two numbers with HCF 1 are called coprime.
R: Two numbers with HCF 1 are called coprime.
Answer: (b). Both statements are true, but R only defines 'coprime'; it is the identity HCF × LCM = ab together with HCF = 1 that explains A. R doesn't directly explain A.
Frequently Asked Questions
What is LCM in Class 7 Maths?
LCM (Lowest Common Multiple) of two or more numbers is the smallest positive number that is a multiple of all of them. LCM(4, 6) = 12. NCERT Class 7 Ganita Prakash Part 2 Chapter 3 explains LCM with listing and prime factorisation.
How do you find LCM by prime factorisation?
Write each number as a product of primes. Take every prime appearing in any factorisation, using the highest power. Multiply these to get LCM. For 4 = 2^2 and 6 = 2 x 3, LCM = 2^2 x 3 = 12. NCERT Class 7 Chapter 3 method.
What is the relationship between HCF and LCM?
For any two positive numbers a and b: HCF(a, b) x LCM(a, b) = a x b. So if you know three of these, you can find the fourth. NCERT Class 7 Ganita Prakash Part 2 Chapter 3 introduces this useful identity.
How to find LCM of 8 and 12?
Prime factorisations: 8 = 2^3, 12 = 2^2 x 3. Highest powers: 2^3 and 3^1. LCM = 2^3 x 3 = 8 x 3 = 24. Verify: 24 is multiple of both 8 and 12. NCERT Class 7 Chapter 3 practices this.
Why is LCM useful in real life?
LCM finds when two periodic events coincide, adds fractions with different denominators, and solves scheduling problems. If bells ring every 4 and 6 minutes, they coincide every 12 minutes (LCM). NCERT Class 7 Chapter 3 gives such examples.
Can LCM be equal to one of the numbers?
Yes, if one number is a multiple of the other, the LCM is the larger number. LCM(4, 8) = 8. In general, LCM(a, b) is at least as large as the larger of a and b. NCERT Class 7 Part 2 Chapter 3 notes this.
Frequently Asked Questions — Chapter 3
What is LCM and Prime Factorisation in NCERT Class 7 Mathematics?
LCM and Prime Factorisation is a key concept covered in NCERT Class 7 Mathematics, Chapter 3: Chapter 3. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on LCM and Prime Factorisation step by step?
To solve problems on LCM and Prime Factorisation, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 3: Chapter 3?
The essential formulas of Chapter 3 (Chapter 3) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is LCM and Prime Factorisation important for the Class 7 board exam?
LCM and Prime Factorisation is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in LCM and Prime Factorisation?
Common mistakes in LCM and Prime Factorisation include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on LCM and Prime Factorisation?
End-of-chapter NCERT exercises for LCM and Prime Factorisation cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.
AI Tutor
Mathematics Class 7 — Ganita Prakash-II
Ready